A Logician’s Toolbox 461: An Introduction to Mathematical Logic Spring 2009 We recast/introduce notions which arise everywhere in mathematics. All proofs are left as exercises. 0 Notations from set theory •{ 1, 2} = {2, 1} = {1, 2, 1} • The ordered pair (a, b) is the set {a, {a, b}}. One has (a, b) = (c, d) iff a = c and b = d. • The Cartesian product of two sets A and B is A × B = {(a, b): a ∈ A, b ∈ B} • The power set of A is the set of all subsets of A, P (A) = {X : X ⊆ A} • When A and B are sets, BA denotes the set of all functions A → B. • 0 ∈ N! 1 Functions Definition 1.1 (function). Function A function from A to B is a relation f on A × B such that: ∀a ∈ A ∀b, b0 ∈ B, (a, b) ∈ f ∧ (a, b0) ∈ f → b = b0 We then write f : A → B, and b = f(a) for (a, b) ∈ f. When talking about a function, it is always very important to make clear its domain dom f and its codomain cod f (A and B respectively in the definition above). To indicate the mapping, the following notation is customary: f : A → B x 7→ f(x) 1 Definition 1.2 (composition). Composition Let f : A → B, g : B → C be functions. Their composition is the function g ◦ f : A → C defined by: (a, c) ∈ g ◦ f if ∃b ∈ B (a, b) ∈ f ∧ (b, c) ∈ g In other words, c = (g ◦ f)(a) iff there is b ∈ B such that b = f(a) and c = g(b) iff c = g(f(a)). Properties 1.3. Composition is associative. If f : A → B, g : B → C, h : C → D are functions, then h ◦ (g ◦ f) = (h ◦ g) ◦ f. 1.1 Injections and Surjections Definition 1.4 (injection). Injection A function f : A → B is injective/an injection if ∀a, a0 ∈ A, f(a) = f(a0) → a = a0 Notation 1.5. Injectivity of f is sometimes denoted f : A,→ B. Properties 1.6. Let f : A → B, g : B → C be functions. • If f and g are injective, so is g ◦ f. • If g ◦ f is injective, so is f. Counter-example 1.7. Let f be the inclusion map from {0} into {0, 1}, and g be the constant map from {0, 1} to {0}. Then g ◦ f is the identity function of {0}, clearly injective, but g isn’t. Definition 1.8 (surjection). Surjection A function f : A → B is surjective/a surjection if ∀b ∈ B ∃a ∈ A, b = f(a) Notation 1.9. Surjectivity of f is sometimes denoted f : A B. Properties 1.10. Let f : A → B, g : B → C be functions. • If f and g are surjective, so is g ◦ f. • If g ◦ f is surjective, so is g. Counter-example 1.11. In counter-example 1.7, g◦f is surjective, but f isn’t. 1.2 Bijections Definition 1.12 (bijection). Bijection A function f : A → B is bijective/a bijection if it is both injective and surjective. Properties 1.13. Let f : A → B, g : B → C be functions. • If f and g are bijective, so is g ◦ f. • If g ◦ f is bijective, then f is injective and g is surjective. Proposition 1.14. Let f : A → B be a bijection. Then there exists a unique function g : B → A such that g ◦ f = IdA and f ◦ g = IdB. Moreover, g is a bijection; it is called the reciprocal bijection of f and denoted f −1. 2 1.3 Image and Pre-image Sets Definition 1.15 (Image set). Image set Let f : A → B be a function and X ⊆ A be a subset of A. The image of X under f is f(X) = {f(x): x ∈ A} Properties 1.16. Let f : A → B be a function and X, Y ⊆ A subsets of A. • f(X ∩ Y ) ⊆ f(X) ∩ f(Y ) • f(X ∪ Y ) = f(X) ∪ f(Y ) Counter-example 1.17. Let f : {0, 1} → {0} be the constant function; set X = {0} and Y = {1}. Then X ∩ Y = ∅ but f(X) ∩ f(Y ) = {0} 6= ∅. Exercise 1.18. Let f : A → B be a function and g defined as follows: g : P (A) → P (B) X 7→ f(X) 1. Show that f is injective iff g is. 2. Show that f is surjective iff g is. Definition 1.19 (Pre-image set). Pre-image Let f : A → B be a function and X ⊆ B be a subset of B. The pre-image of X under f is f −1(X) = {x ∈ A : f(x) ∈ X} Remark 1.20. f −1 as a function from B to A makes sense only when f is a bijection; f −1 as a function from P (B) to P (A) is always defined. It is the case that when f is bijective and X ⊆ B, one has f −1(X) = f −1(X) | {z } | {z } Image under f −1 Pre−image under f Properties 1.21. Let f : A → B be a function and X, Y ⊆ B subsets of B. • f −1(X ∩ Y ) = f −1(X) ∩ f −1(Y ) • f −1(X ∪ Y ) = f −1(X) ∪ f −1(Y ) The following is dual to Exercise 1.18. Exercise 1.22. Let f : A → B be a function and h defined as follows: h : P (B) → P (A) X 7→ f −1(X) 1. Show that f is injective iff h is surjective. 2. Show that f is surjective iff h is injective. 3 1.4 Their interplay Exercise 1.23. Let f : A → B be a function. 1. Let X ⊆ A. Show that f −1(f(X)) ⊆ X. 2. Suppose that f is injective. Show ∀X ⊆ A, f −1(f(X)) = X 3. Suppose ∀X ⊆ A, f −1(f(X)) = X Show that f is injective. Exercise 1.24. Let f : A → B be a function. 1. Let X ⊆ B. Show that X ⊆ f(f −1(X). 2. Suppose that f is surjective. Show ∀X ⊆ B, f(f −1(X)) = X. 3. Suppose ∀X ⊆ B, f(f −1(X)) = X. Show that f is surjective. 2 Equivalence Relations 2.1 Equivalence Relations Definition 2.1 (equivalence relation). Equivalence relation An equivalence re- lation on a set A is a relation ∼⊆ A × A which is reflexive, symmetric, and transitive. We write of course a ∼ b instead of (a, b) ∈∼. Hence the definition is: •∀ a ∈ A, a ∼ a (reflexivity) •∀ a, b ∈ A, a ∼ b → b ∼ a (symmetry) •∀ a, b, c ∈ A, a ∼ b ∧ b ∼ c → a ∼ c (transitivity) Definition 2.2 (equivalence class). Let ∼ be an equivalence relation on a set A. Let a ∈ A. Then the equivalence class of a modulo ∼ is the set [a]∼ = {b ∈ A : a ∼ b} One may drop the subscript ∼ if there is no risk of ambiguity. Properties 2.3. Let ∼ be an equivalence relation on a set A. Let a, b ∈ A. Then a ∼ b iff a ∈ [b]∼ iff b ∈ [a]∼ iff [a]∼ ∩ [b]∼ 6= ∅. Definition 2.4 (quotient set). Quotient set Let ∼ be an equivalence relation on a set A. The quotient set of A by ∼ is the set of all equivalence classes: A/ ∼= {[a]∼ : a ∈ A} 4 2.2 Partitions Definition 2.5 (partition). Partition Let A be a set. A partition of A is a subset Π of P (A) such that: • ∅ 6∈ Π (no member of Π is empty) • S Π = A (members of Π cover A) •∀ X, Y ∈ Π,X ∩Y 6= ∅ → X = Y (distinct members of Π don’t overlap). The typical example is that of the set of classes of an equivalence relation. Remark 2.6. Let ∼ be an equivalence relation on A. Then A/ ∼ is a partition of A. This is no coincidence. Proposition 2.7. Let A be a set. Then there is a bijection between the set of equivalence relations on A and the set of partitions of A. 3 Order Relations 3.1 Pre-orderings Definition 3.1 (pre-ordering). Pre-ordering A pre-ordering on a set A is a relation - which is reflexive and transitive. •∀ a ∈ A, a - a (reflexivity) •∀ a, b, c ∈ A, a - b ∧ b - c → a - c (transitivity) These tend to occur in a mathematician’s everyday life, though they are fairly loose relations. Notice that no one said that elements a and b could always be compared: perhaps neither a - b nor b - a holds. The dullest example of a pre-ordering is the empty relation, that is no two elements are in relation. When a - b ∧ b - a, one says that a and b can be compared. 3.2 Partial Orderings Definition 3.2 (ordering, poset). OrderingPoset An ordering on a set A is a pre-ordering which is anti-symmetric. •∀ a ∈ A, a - a (reflexivity) •∀ a, b ∈ A, a b ∧ b a → a = b (anti-symmetry) •∀ a, b, c ∈ A, a - b ∧ b - c → a - c (transitivity) One then says that (A, ) is a poset. 5 Of course there is a way to retrieve an ordering from any pre-ordering; it suffices to say that elements which both lie above each other must actually be equal. This is done by “factoring out" the natural equivalence relation. Proposition 3.3. Let - be a pre-ordering on a set A. Then • The relation ∼ defined by a ∼ b if a - b ∧ b - a is an equivalence relation on A. • The relation defined by [a]∼ [b]∼ if a - b is an ordering relation on A/ ∼. • The relation defined above is the only order relation on A/ ∼ which is compatible with -.